According to this rule, probabilities of two or more related events are multiplied with each other to find out the net probability of joint occurrence. If A and B are any two events associated with a random experiment, then multiplication rule for A and B is given as: P(A and B) = P(A). = P(B).

Special case of the multiplication rule: If A and B are any two independent events, then multiplication rule for A and B is given as:P(A and B) = P(A ∩ B) = P(A).P(B)

Ex:

An unbiased die is thrown. If A is the event that 'the number appearing is prime' and B be the event that 'the number appearing is more than 4', then find whether A and B are independent ?

Independent events:

Two or more events are said to be independent if occurrence or non-occurrence of any of them does not affect the occurrence or non-occurrence of the other event. For example, when a die is rolled twice, the event of occurrence of '1' in the first throw and the event of occurrence of '1' in the second throw are independent events.

If A and B are any two independent events, then P(A) = and P(B) =

Dependent events:

Two events are said to be dependent if the outcome (or occurrence) of the first effects the outcome(or occurrence) of the second.

3 mutually independent events

Note: Three events A, B, C are said to be mutually independent, if

P(A ∩ B)

=

P(A).P(B)

P(B ∩ C)

=

P(B).P(C)

P(A ∩ C)

=

P(A).P(C) and

P(A ∩ B ∩ C)

=

P(A).P(B).P(C)

If at least one of the above is not true for three given events, we say that the events are not independent.

Some conditional probability problems can be solved by using a tree diagram. A tree diagram is a schematic way of looking at all possible outcomes.

For the case of dependent events, we have seen how to calculate the probability of occurrence of event B if we know that event 'A' has already occurred . Can we calculate the probability of occurrence of 'A' if we know for sure that the event 'B' has occurred? We indicate such probability by P(A/B).
The occurrence of event B happens through two paths as shown below.